Conjugation by g is a homomorphism.  That is, the map x \to gxg^{-1}.  The map x\to gx is a homomorphism if G is assumed to be Abelian.  I'm not certain how the topology of the group would come in to play in this concept, though.

I do not think the map g /to gx is homomorohism when G is abelian. Can you prove it?

Oh, never mind.  I was thinking about the map x\to x^{-1}.  No, that's not a homomorphism.  At least, not in general.

Why you do not use g /to g^{n} for some positive integer $n$?

Dear Aliakbar,

I need a homomorphism which is dependent on a fixed element like x \to g^{-1}xg, but this the trivial hom when G is abelian. Do you have any idea?

Could you give some idea of why you want this homomorphism to depend on g?  As it stands, 'dependence' is too vague to be able to say such a thing might not exist. 

Dear James,

As I explained before like x /to g^-1 xg, but for abelian group. In facr from a sequence of elements in a topological group, I want to construct a sequence of homomorphisms on the group. Do you have any idea?

Let G be a topological  group and fix an element g from G

Now define Fg(x)=gxg-1, then we get Fg(xy)=gxyg-1= gxg-1 gyg-1= Fg(x) Fg(y)

I believe what you are asking is not possible.  Although strictly speaking the following is one solution:  Parametrize the abelian group S^1 by t\in[0,1]. For g = 1/n \in S^1, we have the homomorphism  t \to nt mod 1.  This is a homomorphism of S^1, and depends on g = 1/n.  !  But I feel that is not the kind of answer you are looking for.

This is true but for a very limited example. Anyway, thanks James.

A monoid homomorphism is a semigroup homomorphism that maps the identity element to the identity of the codomain. Suppose G and H are any two groups and f:G to H is a homomorphism, since inverses exist in G and H, one can show that the identity of G maps to the identity of H and that inverses are preserved. Therefore f*g=f*h implies that g=h, g in G and h in H. So g or h is the fixed element.

I have given the answer

What is the structure of your group [0.1]

I think we can not find a law for wihich [0,1] becomes group

@Kahar  He was asking for an abelian group.  Your answer of conjugation becomes the identity homomorphism in that case (and it was suggested several posts earlier).

The example I gave is the circle group S^1, which is R/Z (where [0,1] was just a parametrization).

Ok the circle is commutative group, and the the proof is trivial

We can not work on a set as [0,1], which is not group

we can define a map Fg:G--->G by Fg(x)=gxg-1.

then it will be a homomorphism and also continuous map .